White precipitate of AgCl dissolves in aqueous ammonia solution due to formation of:

Step 1: Notice the reactant and the product characteristics Silver chloride ($\ce{AgCl}$) is a white, curdy precipitate. Aqueous ammonia ($\ce{NH4OH}$) is added to it.

Step 2: Chemical Reaction $\ce{AgCl}$ readily dissolves in aqueous ammonia because it forms a highly stable soluble coordination complex with ammonia ligands. Silver(I), having a $d^{10}$ configuration, preferentially forms linear complexes with a coordination number of 2.

$\ce{AgCl(s) + 2NH3(aq) -> [Ag(NH3)2]Cl(aq)}$

Step 3: Identify the complex The soluble complex specifically formed is Diamminesilver(I) chloride, denoted structurally as $\ce{[Ag(NH3)2]Cl}$.

Conclusion: The solubility is due strictly to the formation of $\ce{[Ag(NH3)2]Cl}$.

Key Points to Remember:

• The silver ion ($\ce{Ag+}$) primarily adopts a coordination number of 2, leading to linear complexes.
• $\ce{AgBr}$ is only sparingly soluble in ammonia, and $\ce{AgI}$ is completely insoluble, providing a key method for separating or differentiating halide precipitates.